Question 85: Electrostatics & Capacitance

Question

3 point charges each of $$-\mathrm{q}$$ are placed on the circumference of a circle of diameter $$2 \mathrm{a}$$ at $$\mathrm{A}, \mathrm{B}$$ and $$\mathrm{C}$$ respectively as shown in figure. The electric field at $$\mathrm{O}$$ is

Accepted Answer

To solve this problem, we need to determine the radii $ R_1, R_2, R_3, $ and $ R_4 $ corresponding to the given potentials and then analyze the relationships provided in the options.
Given:
A uniformly charged solid sphere of radius $ R $.
The potential on its surface is $ V_0 $ (measured with respect to infinity).
Potentials and their corresponding radii:
$ V = \frac{3V_0}{2} $ at $ R_1 $
$ V = V_0 $ at $ R_2 $
$ V = \frac{3V_0}{4} $ at $ R_3 $
$ V = \frac{V_0}{4} $ at $ R_4 $
Electric Potential Due to a Uniformly Charged Solid Sphere:
Outside the Sphere ($ r \geq R $):
$    V(r) = V_0 \cdot \frac{R}{r}    $
Inside the Sphere ($ r \leq R $):
$    V(r) = V_0 \left( \frac{3}{2} - \frac{r^2}{2R^2} ight)    $
Calculations:
Finding $ R_1 $ where $ V = \frac{3V_0}{2} $:
Inside the sphere, at $ r = 0 $:
$      V(0) = V_0 \left( \frac{3}{2} - 0 ight) = \frac{3V_0}{2}      $
Therefore, $ R_1 = 0 $.
Finding $ R_2 $ where $ V = V_0 $:
At the surface ($ r = R $):
$      V(R) = V_0 \left( \frac{3}{2} - \frac{R^2}{2R^2} ight) = V_0      $
Therefore, $ R_2 = R $.
Finding $ R_3 $ where $ V = \frac{3V_0}{4} $:
Outside the sphere ($ r \geq R $):
$      V(r) = V_0 \cdot \frac{R}{r} = \frac{3V_0}{4}      $
$      \frac{R}{r} = \frac{3}{4} \implies r = \frac{4R}{3}      $
Therefore, $ R_3 = \frac{4R}{3} $.
Finding $ R_4 $ where $ V = \frac{V_0}{4} $:
Outside the sphere:
$      V(r) = V_0 \cdot \frac{R}{r} = \frac{V_0}{4}      $
$      \frac{R}{r} = \frac{1}{4} \implies r = 4R      $
Therefore, $ R_4 = 4R $.
Comparisons:
$ R_1 = 0 $
$ R_2 = R $
$ R_3 = \frac{4R}{3} $
$ R_4 = 4R $
Compute $ R_4 - R_3 $:
$ R_4 - R_3 = 4R - \frac{4R}{3} = \left( 4 - \frac{4}{3} ight) R = \frac{8R}{3} $
Compare $ R_2 $ and $ R_4 - R_3 $:
$ R_2 = R, \quad R_4 - R_3 = \frac{8R}{3} \approx 2.67R $
Therefore:
$ R_2 < R_4 - R_3 $
$ R_1 = 0 $
Conclusion:
This matches
Option D
, which states:
$ R_1 = 0 	ext{ and } R_2 < (R_4 - R_3) $
Answer:
$$ \boxed{	ext{Option D: } R_1=0 	ext{ and } R_2<\left(R_4-R_3ight)} $$

Answer

Zero

Answer

$$\frac{\mathrm{Kq}}{2 \mathrm{a}}$$ towards $$\mathrm{OC}$$

Answer

$$\frac{\mathrm{Kq}}{\mathrm{a}}$$ towards $$\mathrm{OB}$$

Answer

$$\frac{\mathrm{Kq}}{\mathrm{a}}$$ towards $$\mathrm{OA}$$

Electrostatics & Capacitance

3 point charges each of $-\mathrm{q} are placed on the circumference of a circle of diameter 2 \mathrm{a} at \mathrm{A}, \mathrm{B} and \mathrm{C} respectively as shown in figure. The electric field at \mathrm{O}$ is

Choose the correct answer: